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{\bf Exam Question

Topic: Double Integral}

Evaluate the integral $$\int_0^2\, dy\int_{2y}^4 \exp(-x^2)\,
dx.$$

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{\bf Solution}

We need to change the order of integration, to give
\begin{eqnarray*}
I&amp;=&amp;\int_0^4\, dx \int_0^{x/2} \exp(-x^2)\, dy =
\int_0^4\frac{1}{2}\exp(-x^2)\, dx\\
&amp;=&amp;\left[-\frac{1}{4}\exp(-x^2)\right]_0^4=\frac{1}{4}\left(1-\mathrm{e}^{-16}\right).
\end{eqnarray*}

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